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Elis [28]
3 years ago
5

Will mark as Brainliest.

Chemistry
2 answers:
kifflom [539]3 years ago
7 0
You’re going to need to use stoichometry to solve this. Hope this helps

postnew [5]3 years ago
3 0
<span>0.453 moles O2 is the solution</span>
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Show the empirical formula manganese fluoride; 59.1% Mn and 40.9% F. Fill in the subscripts on the formula below. Make sure to h
Gre4nikov [31]

Hey there!:

Given % of Mn=59.1% means 59.1 g of Mn present in 100 g of manganese fluoride.

Molar mass of Mn= 54.938 g/mol

Moles of Mn = mass / molar mass

59.1 /54.938 => 1.07 ≈ 1 mol.

and % of F=40.9% means 40.9 g of of F present in 100 g of manganese fluoride.

Molar mass of F=18.998 g/mol

Moles of F :

40.9 / 18.999 => 2.15 mol ≈ 2 mol.

The mole ratio between Mn:F=  1 : 2

Therefore the empirical formula of manganese fluoride:

=> MnF2=Mn1F2

Hope that helps!

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4 years ago
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Answer:

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